Standard error of the slope and the t-test

The slope has a standard error

Eight different clients would have given a slightly different line. ${\hat{\beta }}_1$ is a statistic with a sampling distribution, and under the model of L13.1 it is normal around ${\beta }_1$ with

$$\mathrm{SE}({\hat{\beta }}_1)=\frac{s}{\sqrt{S{S}_{xx}}},S{S}_{xx}=\sum (x-\bar{x}{)}^2=(n-1)s_x^2$$

Two things make the slope precise: a small $s$ (the points hug the line) and a large $S{S}_{xx}$ (the $x$ values are spread out — a slope measured across a wide income range is pinned down better than one measured across a narrow one). Because $s$ replaced $\sigma$, the reference distribution is $t$ on $n-2$ df.

The question the test asks

$H_0:{\beta }_1=0$ says the population line is flat: income carries no linear information about score. $H_a:{\beta }_1\ne 0$ unless the question names a direction. The statistic is Module 10’s shape with the null value zero:

$$t=\frac{{\hat{\beta }}_1-0}{\mathrm{SE}({\hat{\beta }}_1)},df=n-2$$

Worked example — is income linearly related to credit score?

Northfield Credit, $n=8$: ${\hat{\beta }}_1=2.0659$, $s=9.642$, $S{S}_{xx}=4,531.875$. Test at $\alpha =0.01$.

Step 1. $H_0:{\beta }_1=0$, $H_a:{\beta }_1\ne 0$, where ${\beta }_1$ is the change in mean credit score per $1,000 of income among all Northfield clients.

Step 2.

$$\mathrm{SE}({\hat{\beta }}_1)=\frac{9.642}{\sqrt{4,531.875}}=\frac{9.642}{67.32}=0.1432,t=\frac{2.0659}{0.1432}=14.42,df=6$$

Step 3 — critical value. Two-sided at 1 %: $t_{0.005;6}=3.707$. Step 3 — p-value. $14.42$ is beyond the last column of row 6 (3.707 at 0.005), so the one-sided tail is below 0.005 and the two-sided $p<0.01$. The table cannot say more; R will (L13.6: $6.96\times 10^{-6}$).

Step 4. $14.42>3.707$, equivalently $p<0.01$: reject $H_0$ at the 1 % level. There is strong evidence, at the 1 % level, of a linear relationship between income and credit score among Northfield’s clients; each additional $1,000 of income is associated with about 2.07 more points.

What a significant slope does and does not say

It says the slope is distinguishable from zero — a flat line would very rarely produce ${\hat{\beta }}_1=2.07$ from eight points with this scatter. It does not say the relationship is large (that is the slope’s size, in units), tight (that is $R^2$), or causal (these are observational clients; Module 2’s confounders — age, years of credit history — sit behind both variables). Nor does it license predictions outside $35$–$110$: the line was tested where it was fitted.

The intercept row

R prints a $t$ for ${\hat{\beta }}_0$ too ($51.80$). It tests $H_0:{\beta }_0=0$ — whether the line passes through the origin — which is rarely a question anyone asks; the paper wants the slope row.